\(\)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Với mọi \(x\in R\) thì:

\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix} \left|x-1,5\right|=0\\ \left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Khi đó không tồn tại giá trị x

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{1}{6}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\sqrt{\dfrac{1}{6}}\\x=\dfrac{1}{2}-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)